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Let $X$ be a smooth variety over some field $k \subset \mathbb{C}$. By a theorem of Grothendieck, one has a canonical isomorphism of complex vector spaces $$ \mathbb{H}^j(X, \Omega_{X/k}^\bullet) \otimes_k \mathbb{C} \stackrel{\sim}{\longrightarrow} \mathbb{H}^j(X(\mathbb{C}), \Omega^\bullet_{X^{an}/\mathbb{C}}) $$ between algebraic and analytic de Rham cohomology.

The situation with coefficients is more subtle. If $\nabla: E \to E \otimes_k \Omega^1_U$ is some integrable connection on a locally free sheaf $E$ on $X$, there is still a canonical morphism $$ \mathbb{H}^j(X, (E \otimes \Omega_{X/k}^\bullet, \nabla)) \otimes_k \mathbb{C} \longrightarrow \mathbb{H}^j(X(\mathbb{C}), (E^{an} \otimes \Omega^\bullet_{X^{an}/\mathbb{C}}, \nabla^{an})) \quad \quad (\ast) $$ between the algebraic (hyper)cohomology of the complex induced by the connection and the corresponding analytic one.

Deligne proves that this is an isomorphism when $\nabla$ has regular singularities.

Question: is this theorem and "if and only if", that is, if the map (*) is an isomorphism, is it true that $\nabla$ has regular singularities?

If this is the case, a reference would be very much appreciated.

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This should probably be a comment, but it got too long...

Your claim seems to be true if the space $X$ is simply connected. I think it should fail in general, but I am having a hard time coming up with a counterexample.

If $X$ is simply connected, then $(E^{an}, \nabla^{an})$ is isomorphic to the trivial connection. Suppose that

$H^0_{dR}(E,\nabla) \xrightarrow{\sim} H^0_{dR}(E^{an} ,\nabla^{an}) = \mathbb C^r$,

where $r$ is the rank of $E$. Note that $H^0_{dR}(E,\nabla) = Hom_{Conn(X)}((\mathcal O_X,d), (E,\nabla))$, and thus we have a map $f: \mathcal (O_X, d)^{\oplus r} \to (E,\nabla)$ such that the analytification $f^{an}$ is an isomorphism. Hence $f$ is an isomorphism as the analytification functor is conservative. In particular $(E,\nabla) \simeq (\mathcal O_X,d)^{\oplus r}$ is regular.

Thus, in the simply connected case I showed that if analytification induces an isomorphism on $H^0_{dR}$ then the connection is regular.

This stronger statement fails for $X=\mathbb C^\times$. For example, take $E=\mathcal O_X . z^\alpha e^{z}$, where $\alpha \in \mathbb C - \mathbb Z$ (i.e. the trivial line bundle with connection $d + (\alpha/z + 1)dz$). The connection $E$ is non-regular and has no algebraic or analytic flat sections. On the other hand, $H^1_{dR}(E) = \mathbb C$, while $H^1_{dR}(E^{an})=0$ (I think!), so $E$ does not provide a counterexample to your question.

I don't really know much about irregular connections, so I would be curious to see a complete answer to this.

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